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(MATH100)quiz01sol-1b.pdf
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MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Quiz 01 Solutions: Vectors, Vector-valued Functions Duration: 40 minutes Name: ID No.: Tutorial Section: T1B
Answer ALL of the following 4 short answer questions. Show all your work for full credit.
1. (2 Marks) What is the cosine of the angle between the vectors .0, 6, .8. and .1, .1, .1..
Solution: The cosine of the angle, , between any two vectors a and b is de.ned as
cos = a b ,

.a..b. where the numerator is the dot product of a and b and the denominator is the product of the magnitudes of a and b. The dot product of the given vectors is .0, 6, .8..1, .1, .1. = (0)(1) + (6)(.1) + (.8)(.1) = 2. The product of the magnitudes of the given vectors is



..0, 6, .8.. ..1, .1, .1.. = ` 0 + 36 + 64` 1+1+1 = 100 3 =10 3. So that the cosine of the angle between the two vectors is

21 3
= == .
103 53 15
2. (3 Marks) Write down the parametric equations of the line through Q = (0, 3, 4) which is perpen-dicular to both lines
x . 1 y . 2 z . 2 x . 3 z . 1
L1 : == and L2 := ,y =3.
1 .1 .2 .11
Solution: A vector parallel to L1 is .1, .1, .2. and a vector parallel to L2 is ..1, 0, 1.. Thus we can .nd a vector n that is perpendicular to both lines,
.
i jk .

n = .1, .1, .2. ..1, 0, 1. =1 .1 .2 .10 1
.
.1 .2..
1 .2..
1 .1.

= i . j + k = .i + j . k,
01 .11 .10
or
n = ..1, 1, .1..
Hence the symmetric equations of the line through Q and parallel to n is
x . 0 y . 3 z . 4
== ,
.11 .1
or written in the parametric form
x = .t, y =3+ t, z =4 . t.

3. (2 Marks) Let
r(t)= .3 cos t, 3 sin t, 2t. for . <t< .
What is the unit tangent vector to r(t) at t =?
6

Solution: The tangent vector at t = can be found by evaluating the value of the derivative of each
6

component of the position vector at t =:
6

333
..3 sin , 3 cos , 2. = .. ,, 2..
66 22
The magnitude of the above vector is


.
333 .
s.3 .2 .33 .2 r9 27
.. ,, 2.= . + + (2)2 = + +4= 13.
2222 44
Normalize the above vector by dividing each component by its own magnitude, so the unit tangent vector is

3 332
.. , , ..
213213 13
R3
4. (3 Marks) Let C be the curve in de.ned by the parametric equations
x(t) = cos(e t),y(t) = sin(e t),z(t)= e t
fortin [0, 2]. What is the length of C ?

R3
Solution: The length of a curve C in is the integral of the square root of the sum of the squares of the derivatives of each parametric curve over the interval [a, b]. Thus, the derivatives of the above parametric equations are
.tt.tt.t
x (t)= .e sin(e ),y (t)= e cos(e ),z (t)= e.
So that the length of curve C, denoted L, is given by
Z 2 L = p[x.(t)]2 +[y.(t)]2 +[z.(t)]2 dt 0
R 2
= p[.e2t sin(et)]2 +[e2t cos(et)]2 +[et]2 dt
0
Z 2
q
= e2t[cos2(et) + sin2(et)] + e2t dt 0
= Z 2 pe2t + e2t dt = 2 Z 2 e t dt
0 0